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2 years ago

HEELLLPPPPP !

Physics

1 answer:

Sauron [17]2 years ago

4 0

Answer: Alkali metals

Soft, Extremely reactive, 1 valence electron

Alkaline earth metals

Harder, more dense, 2 valence electrons

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Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se

Nady [450]

Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω

7 0

2 years ago

The change in a pitch of a sound as an object moves closer or farther away is a result of

MrMuchimi

The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound.

I’m hope this help

4 0

2 years ago

In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????

34kurt

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

$I = \frac{q}{t}$

<u>Where</u>:

q: is the electric charge transferred through the surface

t: is the time

The charge, q, is:

$q = n*e$

<u>Where</u>:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

$q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C$

Hence, the current in the wire is:

$I = \frac{126.88 C}{3.97 s} = 31.96 A$

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

3 0

2 years ago

Use the information from the article to answer the question.

lbvjy [14]

Answer:

Most asteroids can be found between Mars and Jupiter.

Explanation:

3 0

2 years ago

Read 2 more answers

Consider a series LRC-circuit in which C-120.0 uF. When driven at a frequency w = 200.0 rad s-1 the com ples impedance is given

Lina20 [59]

Answer:

(a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

Explanation:

Given that,

Capacitor = 120.0 μC

Frequency = 200.0 rad/s

Impedance = 100.0 -10j

(I). We need to calculate the $X_{C}$

$X_{C}=\dfrac{1}{C\times\omega}$

Put the value into the formula

$X_{C}=\dfrac{1}{120\times10^{-6}\times200}$

$X_{C}=41.66\ \Omega$

(II). We know that,

Formula of impedance is

$Z=\sqrt{R^2+X_{L}^2+X_{C}^2}$ ...(I)

Given equation of impedance is

$Z=(100-10j)$ ...(II)

On Comparing of equation (I) and (II)

$R = 100$

$X_{L}-X_{C}=-10$

Now, put the value of $X_{C}$

$X_{L=41.66-10$

$X_{L}=31.66\ \Omega$

We need to calculate the inductance

Using formula of inductance

$X_{L}=\omega\times L$

Put the value into the formula

$L=\dfrac{X_{L}}{\omega}$

$L=\dfrac{31.66}{200}$

$L=0.1583\ Henry$

(b). We need to calculate the resonant angular frequency

Using formula of the resonant angular frequency

$angular\ frequency =\dfrac{1}{\sqrt{L\times C}}$

$angular\ frequency =\dfrac{1}{\sqrt{0.1583\times120\times10^{-6}}}$

$angular\ frequency =229.4\ rad/s$

Hence, (a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

4 0

3 years ago